# Contour Integral Question

• Parmenides
In summary, the conversation discusses the integral form encountered when considering the grand potential for a photon gas. The integral is of the form \Sigma = a\int_{0}^{\infty}x^2\ln\Big(1 - e^{-bx}\Big)dx and can be evaluated using contour integration. The individual seeking assistance has not used this method before and is looking for someone to provide a perspective on the contour method for evaluating the integral. It is later recognized that this is a familiar integral found in Stefan's Law, but the individual is still interested in seeing the contour method in action. The contour method involves drawing an imaginary contour in the RE, IM plane, excluding the singularity at x=0, and evaluating

#### Parmenides

When considering the grand potential for a photon gas, one encounters an integral of the form:
$$\Sigma = a\int_{0}^{\infty}x^2\ln\Big(1 - e^{-bx}\Big)dx$$
I have never had to integrate something like this before; I was told it is done via contour integration, but I have never used such a method and the examples typically given are not of this form. Could somebody please provide some assistance? I have tried to learn a bit myself, but I remain perplexed. What would be the contour enclosing such an integral, for example? Thanks.

UPDATE: I noticed that integration by parts puts the integral in a form of $$\int_{0}^{\infty}\frac{x^3dx}{e^{bx} - 1}$$ ignoring constants. I now recognize this as a familiar integral found in Stefan's Law, but it would still be nice to see someone's perspective of the contour method to evaluate it.

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Looking at your updated integral, this is an analytic function away from the singularity (simple pole) at x=0. The contour method would have you draw an imaginary contour in the RE, IM plane which traces the imaginary axis from +infty to +epsilon*i then makes a small half circle cut to +epsilon and follows the real axis to +infty, then has a semicircular closure with radius +infty. This contour excludes the singularity at 0, so clearly is analytic on the interior of the contour, which means it will be zero.

Therefore, the evaluation of your integral simplifies to the evaluation of the residue at x=zero.